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Question 1 : An airline wants to select a computer software package for its rese

ID: 3154657 • Letter: Q

Question

Question 1: An airline wants to select a computer software package for its reservation system. Four software packages (1, 2, 3, and 4) are commercially available. The airline will choose the package that bumps as few passengers as possible during a month. An experiment is set up in which each package is used to make reservations for 5 randomly selected weeks. (A total of 20 weeks was included in the experiment.) The number of passengers bumped each week is obtained, which gives rise to the following Excel output:

ANOVA   

Source of Variation          SS            df             MS           F                  P-value        F crit

Between Groups              212.4       3                              8.304985      0.001474     3.238867     

Within Groups                 136.4                       8.525                                               

Total                                348.8   

  

a) What is the within groups degrees of freedom for this analysis? Explain how you obtain your answer.

b) What is the total degrees of freedom? Explain how you obtain your answer.

c) What is the among-group (between-group) mean squares? Explain how you obtain your answer.

d) At a significance level of 1%, what conclusion can you infer? Explain how you obtain your answer.

Explanation / Answer

a) The within groups degrees of freedom for this analysis is 16

b) The total degrees of freedom for this analysis is 19

Expalnation for a and b: The total number of weeks n = 20

Hence total degrees of freedom = (n-1) = (20-1) = 19

Since there 4 packages available the between groups degrees of freedom = (4-1) = 3

Hence

The within groups degrees of freedom = Total degrees of freedom - Between groups degrees of freedom = 19-3 =16

c) The among-group (between-group) mean squares is given by SS/df = 212.4/3 = 70.8

d) Since the p-value is 0.001474 < 0.01(significance level), we can infer that all the means are not equal at 1% significance level.

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